S4 DAT Practice Paper (Set 2): Structures And Moments, Gear Ratio, Material Density And Stress, Series Circuits And Power, Tolerance And Energy Efficiency

S4 DAT Practice Paper (Set 2): Structures and Moments, Gear Ratio, Material Density and Stress, Series Circuits and Power, Tolerance and Energy Efficiency

2026-11-15

S4 DAT

S4 DAT paper (Set 2) with answers: moments and reactions, second moment of area, gear ratio, material density, tensile strength, series circuits and energy efficiency.

Paper details　S4 DAT practice paper (Set 2) with full answers covering moments, gear ratio, material density, tensile strength, series circuits, tolerance and energy efficiency.

Structures and moments: cantilever load, reaction and section

A 2.4 m aluminium beam is fixed at its left end. A load of 150 N hangs 1.8 m from the support. The section measures 40 mm by 60 mm and the allowable bending moment of the material is 810 N m.

#	Question	Answer and marking notes
1	Calculate the moment of the load about the support.	Answer: Moment = load x lever arm = 150 N x 1.8 m = 270.0 N m.
2	A prop is added 0.6 m from the support. Find the upward reaction it must carry.	Answer: Taking moments about the support, reaction x 0.6 m = 270.0 N m, so the reaction is 450.0 N.
3	Compute the second moment of area with 40 mm width and 60 mm depth, and with 60 mm width and 40 mm depth, then give the ratio.	Answer: I = b x h^3 / 12. The first is 720000 mm4 and the second is 320000 mm4, a ratio of 2.25 times.
4	Find the factor of safety of the beam under the design load.	Answer: Factor of safety = allowable moment / working moment = 810 N m / 270.0 N m = 3.0.

Gear train: ratio, speed and output power

A motor runs at 1440 rpm on an 18-tooth driving gear that drives a 54-tooth driven gear. The transmission efficiency is 0.90 and the input torque is 2.4 N m.

#	Question	Answer and marking notes
5	Find the gear ratio.	Answer: Gear ratio = driven teeth / driving teeth = 54 / 18 = 3.0.
6	Find the speed of the driven shaft.	Answer: Driven speed = 1440 rpm / 3.0 = 480.0 rpm.
7	Find the output torque.	Answer: Output torque = input torque x ratio x efficiency = 2.4 N m x 3.0 x 0.90 = 6.48 N m.
8	Find the output power.	Answer: Power = 2 pi x speed x torque / 60 = 2 pi x 480.0 x 6.48 / 60 = 325.7 W.

Materials and stress: volume, mass, cost and tensile strength

A PET sheet measures 300 mm by 200 mm by 5 mm, density 1.38 g/cm3, priced at 24 dollars per kilogram. A tensile specimen of section 20 mm by 5 mm breaks at 3500 N.

#	Question	Answer and marking notes
9	Find the volume of the sheet in cm3.	Answer: Volume = 30 cm x 20 cm x 0.5 cm = 300.0 cm3.
10	Find the mass of the sheet in grams.	Answer: Mass = volume x density = 300.0 cm3 x 1.38 g/cm3 = 414.0 g.
11	Find the material cost of the sheet.	Answer: Mass is 0.414 kg, so cost = 0.414 kg x 24 dollars = 9.94 dollars.
12	Find the tensile strength in N/mm2.	Answer: Tensile strength = breaking load / area = 3500 N / 100 mm2 = 35.0 N/mm2.

Series circuit: total resistance, current, voltage division and power

A 12 V d.c. supply is connected in series with a 220 ohm resistor and a 330 ohm resistor.

#	Question	Answer and marking notes
13	Find the total resistance of the circuit.	Answer: Total resistance = 220 ohm + 330 ohm = 550 ohm.
14	Find the circuit current in mA.	Answer: Current = voltage / total resistance = 12 V / 550 ohm = 0.02182 A, that is 21.8 mA.
15	Find the voltage across the 330 ohm resistor.	Answer: Voltage = 12 V x 330 ohm / 550 ohm = 7.2 V.
16	Find the power dissipated in the 330 ohm resistor in mW.	Answer: Power = voltage squared / resistance = 7.2^2 / 330 ohm = 0.15709 W, that is 157.1 mW.

Tolerance and dimension chains: fits and accumulated tolerance

A hole and shaft fit is specified as: hole 20.00 to 20.05 mm, shaft 19.91 to 19.96 mm. A dimension chain is built from three parts of 45.0 plus or minus 0.2, 30.0 plus or minus 0.1 and 12.0 plus or minus 0.1.

#	Question	Answer and marking notes
17	State the maximum and minimum hole sizes.	Answer: The maximum hole is 20.05 mm and the minimum hole is 20.0 mm.
18	State the maximum and minimum shaft sizes.	Answer: The maximum shaft is 19.96 mm and the minimum shaft is 19.91 mm.
19	Find the maximum and minimum clearance.	Answer: Maximum clearance = 20.05 – 19.91 = 0.14 mm; minimum clearance = 20.0 – 19.96 = 0.04 mm.
20	Find the total length and total tolerance of the dimension chain.	Answer: Total length = 45.0 + 30.0 + 12.0 = 87.0 mm; total tolerance = 0.2 + 0.1 + 0.1 = plus or minus 0.4 mm.

Energy efficiency and cost: lighting replacement and motor losses

A 12 W LED fitting replaces a 60 W filament lamp and runs 5 hours a day. Electricity costs 1.2 dollars per kWh. A motor takes 200 W of input power and delivers 150 W of mechanical power.

#	Question	Answer and marking notes
21	Find the energy saved each day in kWh.	Answer: Daily saving = (60 W – 12 W) x 5 h = 240 Wh, that is 0.24 kWh.
22	Find the energy saved in a year of 365 days.	Answer: Yearly saving = 0.24 kWh x 365 = 87.6 kWh.
23	Find the money saved each year.	Answer: Money saved = 87.6 kWh x 1.2 dollars = 105.12 dollars.
24	Find the motor efficiency and the energy lost when 2.4 kWh is supplied.	Answer: Efficiency = 150 W / 200 W = 0.75, that is 75 per cent; energy lost = 2.4 kWh x (1 – 0.75) = 0.6 kWh.

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